Data Communication and Networking - Chapter 1: Transmission Media and PHY Layer - Vo Que Son

Nội dung text: Data Communication and Networking - Chapter 1: Transmission Media and PHY Layer - Vo Que Son

1. STP - Shield Twisted Pair Intrinsic Impedance: usually 150 Ohm Speed: 10,100 Mbps depending on cable quality/category Average $ per node: moderately expensive Maximum cable length: 100m Media and connector size: medium to large Telecomm. Dept. DCN 9 Faculty of EEE HCMUT
2. Most of common cables Name Type Bandwidth Applications Level 1 0.4 MHz Telephone and modem lines Level 2 4 MHz Older terminal systems, e.g. IBM 3270 Cat3 UTP 16 MHz 10BASE-T and 100BASE-T4 Ethernet Cat4 UTP 20 MHz 16 Mbit/s Token Ring Cat5 UTP 100 MHz 100BASE-TX & 1000BASE-T Ethernet Cat5e UTP 100 MHz 100BASE-TX & 1000BASE-T Ethernet Cat6 UTP 250 MHz 10GBASE-T Ethernet Cat6a 500 MHz 10GBASE-T Ethernet Telephone, CCTV, 1000BASE-TX in the Class F S/FTP 600 MHz same cable. 10GBASE-T Ethernet. Telephone, CATV, 1000BASE-TX in the same Class Fa 1000 MHz cable. 10GBASE-T Ethernet. Telecomm. Dept. DCN 11 Faculty of EEE HCMUT
3. Cabling Types  Switch to router  Switch to PC or server  Hub to PC or server  Switch to switch  Switch to hub  Hub to hub  Router to router  PC to PC  Router to PC Telecomm. Dept. DCN 13 Faculty of EEE HCMUT
4. Co-axial Cable (1)  Applications: Computer Networks, Data Systems, CATV, Private Video Networks  RG-6/RG-59: 75 Ohm, CATV  RG-8/ RG-58: 50 Ohm, Thick and Think Ethernet LANs  RG-62: 93 Ohm, used in IBM Mainframe EMI prevention Many kinds of intrinsic impedances: co-axial cables can be used in specific systems Telecomm. Dept. DCN 15 Faculty of EEE HCMUT
5. Reflection and Refraction  Electromagnetic waves pointing out from the source are called rays  When a ray of light (the incident ray) strikes the shiny surface of a flat piece of glass, some of the light energy in the ray is reflected Telecomm. Dept. DCN 17 Faculty of EEE HCMUT
6. Multi-mode fiber Subscriber Connector (SC) is used in multimode Telecomm. Dept. DCN 19 Faculty of EEE HCMUT
7. Attenuation Telecomm. Dept. DCN 21 Faculty of EEE HCMUT
8. Fiber: characteristics  Fiber-optic cable is not affected by the sources of external noise  Fiber in a cable does not generate interference that disturbs transmission on any other fiber  The scattering of light, Absorption, Dispersion, manufacturing irregularities  Expensive and complicated installation Telecomm. Dept. DCN 23 Faculty of EEE HCMUT
9. Cabling Evolution (1)  Telephony PABX Telecomm. Dept. DCN 25 Faculty of EEE HCMUT
10. Cabling Evolution (3) IEEE 802.3 Ethernet 1980 10 BASE-5 SAS 10 BASE-2 10 BASE-Tx 10 BASE-FL TP-PMD FD DI 10BASE-5 DAS Coxial FDDI 10BASE-2 Fiber Distributed Data Interface 1980s IEEE 802.5 Token Ring Token Ring 4 mbps 1985 CAT3 UTP Token Ring 16 mbps 1989 IBM Type1 Token Ring 2-pairSTP ATM Telecomm. Dept. 1-27 DCN 27 Faculty of EEE HCMUT
11. Cabling Evolution (3) Universal Cabling Two media: copper twisted pair and optical fiber Voice, data, video, control signals Patching Facilities Universal (Generic): application independent Telecomm. Dept. DCN 29 Faculty of EEE HCMUT
12. Benefits of Structured Cabling Redundancy at design stage reduces downtime & repair time Concealed cabling Ease of fault location & repair Flexibility, expandability & modular connecting platform Ease of moves, adds and changes Enhanced end-use understanding and control Continuous product support and warranty Significant long term cost containment Telecomm. Dept. DCN 31 Faculty of EEE HCMUT
13. Campus Distributor < 1500 m Campus Distributor < 1500 m < 1500 m Telecomm. Dept. DCN 33 Faculty of EEE HCMUT
14. Elements of a Structured Cabling System Telecomm. Dept. DCN 35 Faculty of EEE HCMUT
15. Wireless media Satellite Microwave Terrestrial Microwave Infrared Telecomm. Dept. DCN 37 Faculty of EEE HCMUT
16. Satellite systems Telecomm. Dept. DCN 39 Faculty of EEE HCMUT
17. Satellite frequency bands Applications: . Broadcast, TV . Long-haul telephone system . Private business network Telecomm. Dept. DCN 41 Faculty of EEE HCMUT
18. Broadcast Radio Radio is 3kHz to 300GHz use broadcast radio, 30MHz - 1GHz, for: FM radio UHF and VHF television is omnidirectional still need line of sight suffers from multipath interference reflections from land, water, other objects Telecomm. Dept. DCN 43 Faculty of EEE HCMUT
19. Terrestrial Microwave Propagation model Telecomm. Dept. DCN 45 Faculty of EEE HCMUT
20. Terrestrial Microwave Applications Long-haul voice communications Common carriers Private network Characteristics: Using sky-wave Line-of-sight Frequency band: 2 – 40GHz Sensitive to obstacles, environment changes. Telecomm. Dept. DCN 47 Faculty of EEE HCMUT
21. Infrared Infrared (IR) light is electromagnetic radiation with longer wavelengths than those of visible light Light-of-sight Applications: PC-PC, PDA communication Data communication in small networks Telecomm. Dept. DCN 49 Faculty of EEE HCMUT
22. RS232  RS-232 is a popular communications interface for connecting modems and data acquisition devices (i.e. GPS receivers, electronic balances, data loggers, ) to computers.  RS-232 can be plugged straight into the computer’s serial port (know as COM or Comm port).  Components of standard:  Connection must be less than 50 feet  Data represented by voltages between +15v and -15v  25-pin connector, with specific signals such as data, ground and control assigned to designated pins  Specifies transmission of characters between, e.g., a terminal and a modem  Transmitter never leaves wire at 0v; when idle, transmitter puts negative voltage (a 1) on the wire Telecomm. Dept. DCN 51 Faculty of EEE HCMUT
23. RS232 Signal  Architecturally RS-232 is a bi- directional point to point link.  Two independent channels are (serial port - PC side) established for two-way (full- duplex) communications.  RS-232 can also carry additional signals used for flow control (RTS, CTS) and modem control (DCD, DTR, DSR, RI). Telecomm. Dept. DCN 53 Faculty of EEE HCMUT
24. RS232 Line Driver  Unbalanced Line Drivers  Each signal appears on the interface connector as a voltage with reference to a signal ground.  The “idle” state (MARK) has the signal level negative with respect to common whereas the active state (SPACE) has the signal level positive respect to the same reference. Telecomm. Dept. DCN 55 Faculty of EEE HCMUT
25. DTE-DCE connection Telecomm. Dept. DCN 57 Faculty of EEE HCMUT
26. RS232: Null modem  Null modem is a communication method to connect two DTEs (computer, terminal, printer etc.) directly using an RS-232 serial cable.  Null modems were commonly used for file transfer between computers, or remote operation  Types:  No hardware handshaking  Loop back handshaking  Partial handshaking Telecomm. Dept. DCN 59 Faculty of EEE HCMUT
27. RS422 Signal Bit representation: Bit 1: +V and –V Bit 0: -V and +V Details: Physical media: twisted-pair Topology: P2P, Multi-dropped Voltage Levels −6V to +6V +6V (maximum differential Voltage) -6V Maximum Drivers: 10 (1 driver, 10 receivers) Bit rate: 100 kbps – 10 Mbps Telecomm. Dept. DCN 61 Faculty of EEE HCMUT
28. RS422: extend RS232 Telecomm. Dept. DCN 63 Faculty of EEE HCMUT
29. RS485 Line Driver  Bus topology  Point-to-Point  Multi-drop  Multi-point  Slave devices: 255  Physical media: Balanced Interconnecting Cable  Full-duplex: 4 wires  Half-duplex: 2 wires Telecomm. Dept. DCN 65 Faculty of EEE HCMUT
30. RS485 Network  RS-485 provides Half- Duplex, Multidrop communications over a single twisted pair cable.  The standard specifies up to 32 drivers and 32 receivers can share a multidrop network  Terminator resistors avoid reflected signal Telecomm. Dept. DCN 67 Faculty of EEE HCMUT
31. RS485 hub Expand network Telecomm. Dept. DCN 69 Faculty of EEE HCMUT
32. Terms Telecomm. Dept. DCN 71 Faculty of EEE HCMUT
33. Effect of lack of synchronization  In a digital transmission, the receiver clock is 0.1 percent faster than the sender clock. How many extra bits per second does the receiver receive if the data rate is 1 kbps? How many if the data rate is 1 Mbps?  Solution - At 1 kbps, the receiver receives 1001 bps instead of 1000 bps. - At 1 Mbps, the receiver receives 1,001,000 bps instead of 1,000,000 bps. Telecomm. Dept. DCN 73 Faculty of EEE HCMUT
34. Unipolar NRZ Non Return to Zero Bit 1: positive voltage, bit 0: 0V Constant voltage in bit duration Lack of clock synchronization Having DC component Easy for deployment Telecomm. Dept. DCN 75 Faculty of EEE HCMUT
35. Polar NRZ  In NRZ-L the level of the voltage determines the value of the bit. In NRZ-I the inversion or the lack of inversion determines the value of the bit.  NRZ-L and NRZ-I both have an average signal rate of N/2 data rate  Advantages:  Simple implementation  Efficiency of using bandwidth  Disadvantages:  Having DC component  Lack of synchronization  Applications: in magnetic recorder, not used in transmission Telecomm. Dept. DCN 77 Faculty of EEE HCMUT
36. Bi-phase In Manchester and differential Manchester encoding, the transition at the middle of the bit is used for synchronization. The minimum bandwidth of Manchester and differential Manchester is 2 times that of NRZ. Manchester: bit 1 –V +V, bit 0: +V -V Differential Manchester: Same as Manchester Next bit is bit 1: no inversion of voltage Next bit is bit 0: inversion of voltage Telecomm. Dept. DCN 79 Faculty of EEE HCMUT
37. Bi-phase Advantages: Synchronization: using clock-edge at the middle of bit No DC component Error detection: unexpected clock-edge happens Disadvantages: Bandwidth: 2 times of NRZ Telecomm. Dept. DCN 81 Faculty of EEE HCMUT
38. mBnL scheme In mBnL schemes, a pattern of m data elements is encoded as a pattern of n signal elements in which 2m ≤ Ln. Increase data rate Decrease required bandwidth Definitions: m: length of binary pattern B: binary data n: length of signal pattern L: number of levels in the signaling Telecomm. Dept. DCN 83 Faculty of EEE HCMUT
39. Multilevel: 8B6T scheme  Number of binary patterns: 28=256  Number of signal level: 36=478  Redundant: 222  Clock synchronization  Error detection  DC balance Telecomm. Dept. DCN 85 Faculty of EEE HCMUT
40. Block coding Example of block coding: 4B5B 8B10B Telecomm. Dept. DCN 87 Faculty of EEE HCMUT
41. B8ZS  Bipolar with 8-Zero Substitution (B8ZS)  Replace 8 consecutive zeros by 000VB0VB  V: denoted for violation: breaks the AMI rule (opposite polarity from the previous)  B: denoted for bipolar (follows the AMI rule) Telecomm. Dept. DCN 89 Faculty of EEE HCMUT
42. Digital Modulation/Demodulation  Digital-to-analog conversion is the process of changing one of the characteristics of an analog signal based on the information in digital data.  Types: Telecomm. Dept. DCN 91 Faculty of EEE HCMUT
43. Example An analog signal carries 4 bits per signal element. If 1000 signal elements are sent per second, find the bit rate (c=1) Solution: In this case, r = 4, S = 1000, and N is unknown. We can find the value of N from Telecomm. Dept. DCN 93 Faculty of EEE HCMUT
44. Binary amplitude shift keying Waveform: Simple implementation Used in telematics Telecomm. Dept. DCN 95 Faculty of EEE HCMUT
45. Binary amplitude shift keying Implementation of Binary ASK Telecomm. Dept. DCN 97 Faculty of EEE HCMUT
46. Binary ASK: full-duplex  In data communications, we normally use full-duplex links with communication in both directions. We need to divide the bandwidth into two with two carrier frequencies, as shown in Figure 5.5. The figure shows the positions of two carrier frequencies and the bandwidths. The available bandwidth for each direction is now 50 kHz, which leaves us with a data rate of 25 kbps in each direction. Telecomm. Dept. DCN 99 Faculty of EEE HCMUT
47. Binary frequency shift keying  More complicated than ASK  Difficult in synchronization  Used in Data Communication Modems (CCITT V21, CCITT V23, BELL 103, BELL 113, BELL 202) and digital radio  Lower BER than ASK: why? Telecomm. Dept. DCN 101 Faculty of EEE HCMUT
48. Binary frequency shift keying  BFSK demodulation using PLL  BFSK coherent demodulation  BFSK non-coherent demodulation Telecomm. Dept. DCN 103 Faculty of EEE HCMUT
49. Multi-level FSK  We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth.  Solution: We can have L = 23 = 8. The baud rate is S = 3 Mbps/3 = 1 MHz. This means that the carrier frequencies must be 1 MHz apart (2Δf = 1 MHz). The bandwidth is B = 8 × 1000 = 8000. Figure 5.8 shows the allocation of frequencies and bandwidth. Telecomm. Dept. DCN 105 Faculty of EEE HCMUT
50. Binary phase shift keying  Medium complex circuit  Used in digital radio  BPSK implementation: Telecomm. Dept. DCN 107 Faculty of EEE HCMUT
51. QPSK Demodulation scheme: Telecomm. Dept. DCN 109 Faculty of EEE HCMUT
52. Quadrature Amplitude Modulation Phase and Amplitude Increase bit rate Telecomm. Dept. DCN 111 Faculty of EEE HCMUT
53. Delay Example A data frame of 1000 bits is transmitted between 2 DTE. Determine RTT depending on which kind of delay (Tp or Tx) in the following cases: 2 DTEs are connected by twisted-pair cable with a distance of 100m, R=10 Kbps 2 DTEs are connected by co-axial cable with a distance of 10km, R=1 Mbps 2 DTEs are connected via free space with a distance of 5000km, R=10 Mbps Given v=2.108 m/s Telecomm. Dept. DCN 113 Faculty of EEE HCMUT
54. Baseband transmission  An example of a dedicated channel where the entire bandwidth of the medium is used as one single channel is a LAN. Almost every wired LAN today uses a dedicated channel for two stations communicating with each other. In a bus topology LAN with multipoint connections, only two stations can communicate with each other at each moment in time (timesharing); the other stations need to refrain from sending data. In a star topology LAN, the entire channel between each station and the hub is used for communication between these two entities Telecomm. Dept. DCN 115 Faculty of EEE HCMUT
55. Bandpass channel  An example of broadband transmission using modulation is the sending of computer data through a telephone subscriber line, the line connecting a resident to the central telephone office. These lines are designed to carry voice with a limited bandwidth. The channel is considered a bandpass channel. We convert the digital signal from the computer to an analog signal, and send the analog signal. We can install two converters to change the digital signal to analog and vice versa at the receiving end. The converter, in this case, is called a modem Telecomm. Dept. DCN 117 Faculty of EEE HCMUT
56. Attenuation  Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as  A loss of 3 dB (–3 dB) is equivalent to losing one-half the power. Telecomm. Dept. DCN 119 Faculty of EEE HCMUT
57. Example  The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?  Solution: The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as Telecomm. Dept. DCN 121 Faculty of EEE HCMUT
58. SNR  Two cases of SNR: a high SNR and a low SNR Telecomm. Dept. DCN 123 Faculty of EEE HCMUT
59. Example  We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?  Solution: We can use the Nyquist formula as shown:  Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps. Telecomm. Dept. DCN 125 Faculty of EEE HCMUT
60. Example  We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000. The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as  This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise ratio. Telecomm. Dept. DCN 127 Faculty of EEE HCMUT
61. Bandwidth-Delay product  In networking, we use the term bandwidth in two contexts.  The first, bandwidth in hertz, refers to the range of frequencies in a composite signal or the range of frequencies that a channel can pass.  The second, bandwidth in bits per second, refers to the speed of bit transmission in a channel or link. Telecomm. Dept. DCN 129 Faculty of EEE HCMUT
62. Gaussian noise and BER  Noise: always present in the transmission environment  Modeled by Gaussian distribution  Effected to the signal quality at the receiver Telecomm. Dept. DCN 131 Faculty of EEE HCMUT
63. Gaussian noise and BER  Power Density Function (pdf) of Gaussian distribution: m: mean (DC value) : Standard Deviation (RMS/Effective Voltage) 2  : variance (noise power) p(x) 1 2  2 0.606   xm 2 2 1 2 2  p xe 2 2 2  0.136 2 2 2  2 0 m x Telecomm. Dept. DCN 133 Faculty of EEE HCMUT
64. Gaussian noise and BER  Probability of transmitting bit 1, but receiving bit 0: 2 vT (x A) 1 2 P (v v ) = P (1/0) = e 2 dx r D T r 2 vT 2   Assuming that the probability of transmitting bit 1 and bit 0 is the same: Pr(1)=Pr(0)=0.5 Telecomm. Dept. DCN 135 Faculty of EEE HCMUT
65. Gaussian noise and BER Finally, we need to determine Pr(1/0) or Pr(0/1): using Q(k) function Q(k) is a normal distribution function m=0, =1 In this case: VT=A/2(why?) VT/=A/2 so: Area is Q(k) Pe=Q(A/2) Let k= A/2: Pe=Q(k) Telecomm. Dept. DCN 137 Faculty of EEE HCMUT
66. Frame Error Rate  Given Pe  If a frame has n bits. What is the Frame Error Rate?  Probability of k-bit error: kkn-k P=CP(1-P)knee  P0 is the probability that there is no errors in the frame  Frame Error Rate (FER): n Pf(error) = 1-P0 = 1-(1-Pe ) nPe (because Pe<<1)  What is the probability of one frame has exact k-bit continuous burst error? Telecomm. Dept. DCN 139 Faculty of EEE HCMUT
67. Example: Optimum Threshold Voltage Let p1 the transmission probability of symbol –A, and p2 is the one of symbol A. Determine the optimum value of threshold voltage VT. Assuming the noise is Gaussian noise with variance N0? Solution: Telecomm. Dept. DCN 141 Faculty of EEE HCMUT